Dehydration of 2‑methyl‑3‑butanol ($\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_3$) in conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ at $170^{c}ircmathrmC$ gives a major alkene product. Identify the product and explain the reasoning.

Step-by-step solution

1. Step 1: Identify the carbon bearing the OH group – it is a secondary alcohol at C3.
2. Step 2: Dehydration removes water to form a double bond. Two possible alkenes: 2‑methyl‑$2$‑butene (double bond between C2–C3) or 2‑methyl‑$3$‑butene (double bond between C3–C4).
3. Step 3: Count β‑hydrogens: both C2 and C4 have two β‑hydrogens each – equal number of possibilities.
4. Step 4: Apply Zaitsev’s rule: the more substituted alkene is favoured. 2‑methyl‑$2$‑butene is trisubstituted, whereas 2‑methyl‑$3$‑butene is disubstituted.
5. Step 5: Conclude that 2‑methyl‑$2$‑butene is the major product.

Concept tested: Dehydration of secondary alcohols – Zaitsev’s rule

Published 11 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked